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COMPSCI 102

COMPSCI 102. Introduction to Discrete Mathematics. Counting I: One-To-One Correspondence and Choice Trees. Lecture 6 (September 15, 2010). If I have 14 teeth on the top and 12 teeth on the bottom, how many teeth do I have in all?. Addition Rule.

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COMPSCI 102

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  1. COMPSCI 102 Introduction to Discrete Mathematics

  2. Counting I: One-To-One Correspondence and Choice Trees Lecture 6 (September 15, 2010)

  3. If I have 14 teeth on the top and 12 teeth on the bottom, how many teeth do I have in all?

  4. Addition Rule Let A and B be two disjoint finite sets The size of (A  B) is the sum of the size of A and the size of B

  5. Addition Rule (2 possibly overlapping sets) Let A and B be two finite sets |AB| = |A| + |B| - |AB|

  6. Addition of multiple disjoint sets: • Let A1, A2, A3, …, An be disjoint, finite sets.

  7. Partition Method • To count the elements of a finite set S, partition the elements into • non-overlapping subsets A1, A2, A3, …, An .. • |s| =

  8. Partition Method S = all possible outcomes of one white die and one black die.

  9. Partition Method S = all possible outcomes of one white die and one black die. Partition S into 6 sets: • A1 = the set of outcomes where the white die is 1. • A2 = the set of outcomes where the white die is 2. • A3 = the set of outcomes where the white die is 3. A4 = the set of outcomes where the white die is 4. A5 = the set of outcomes where the white die is 5. A6 = the set of outcomes where the white die is 6. Each of 6 disjoint sets have size 6 = 36 outcomes

  10. Partition Method S = all possible outcomes where the white die and the black die have different values

  11. S  Set of all outcomes where the dice show different values. S = ? • Ai set of outcomes where black die says i and the white die says something else.

  12. S  Set of all outcomes where the dice show different values. S = ? T  set of outcomes where dice agree. = { <1,1>, <2,2>, <3,3>,<4,4>,<5,5>,<6,6>} | S  T | = # of outcomes = 36 |S| + |T| = 36 |T| = 6 |S| = 36 – 6 = 30

  13. How many seats in this auditorium? Count without Counting: The auditorium can be Partitioned into n rows with k seats each Thus, we have nk seat in the room

  14. S  Set of all outcomes where the black die shows a smaller number than the white die. S = ? Ai set of outcomes where the black die says i and the white die says something larger. S = A1 A2 A3 A4 A5 A6 |S| = 5 + 4 + 3 + 2 + 1 + 0 = 15

  15. S  Set of all outcomes where the black die shows a smaller number than the white die. S = ? L  set of all outcomes where the black die shows a larger number than the white die. S + L = 30 It is clear by symmetry that | S | = | L |. Therefore | S | = 15

  16. “It is clear by symmetry that |S| = |L|?”

  17. Pinning Down the Idea of Symmetry by Exhibiting a Correspondence Put each outcome in S in correspondence with an outcome in L by swappingcolor of the dice. S L Each outcome in S gets matched with exactly one outcome in L, with none left over. Thus:S = L

  18. Let f : A  B Be a Function From a Set A to a Set B • f is 1-1 if and only if • x,yÎA, x ¹ y Þ f(x) ¹ f(y) f is onto if and only if zÎB xÎA f(x) = z There Exists For Every

  19. A B Let’s Restrict Our Attention to Finite Sets  1-1 f : AB | A | ≤ | B |

  20. B A  onto f : AB | A | ≥ | B |

  21. A B  1-1 and onto f : A B | A | = | B |

  22. A B f is1-1 and onto means f-1 is uniquef is a way of pairing up elements

  23. Correspondence Principle If two finite sets can be placed into 1-1 onto correspondence, then they have the same size It’s one of the most important mathematical ideas of all time!

  24. 0 000000  1 000001  2 000010  3 000011 : : : 111111  2n-1 Question: How many n-bit sequences are there? Each sequence corresponds to a uniquenumber from 0 to 2n-1. Hence 2n sequences.

  25. A = { a,b,c,d,e } Has Many Subsets {a}, {a,b}, {a,d,e}, {a,b,c,d,e}, {e}, Ø, … The entire set and the empty set are subsets with all the rights and privileges pertaining thereto

  26. Question: How Many Subsets Can Be Made From The Elements of a 5-Element Set? a b c d e 0 1 1 0 1 1 means “TAKE IT” { b c e } 0 means “LEAVE IT” Each subset corresponds to a 5-bit sequence (using the “take it or leave it” code)

  27. a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 A = {a1, a2, a3,…, an}B = set of all n-bit strings For bit string b = b1b2b3…bn, let f(b) = { ai | bi=1} Claim: f is 1-1 Any two distinct binary sequences b and b have a position i at which they differ Hence, f(b) is not equal to f(b) because they disagree on element ai

  28. a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 A = {a1, a2, a3,…, an}B = set of all n-bit strings • Let S be a subset of {a1,…,an}. • Define bk = 1 if ak in S and bk = 0 otherwise. • Note that f(b1b2…bn) = S. For bit string b = b1b2b3…bn, let f(b) = { ai | bi=1} Claim: f is onto

  29. The number of subsets of an n-element set is 2n

  30. Let f : A  B Be a Function From Set A to Set B f is 1-1 if and only if x,y A, x  y  f(x)  f(y) f is onto if and only if zB xA such that f(x) = z

  31. A B Let f : A  B Be a Function From Set A to Set B f is a 1 to 1correspondence iff zB  exactly one xA such that f(x) = z f is a k to 1correspondence iff zB  exactly k xA such that f(x) = z 3 to 1 function

  32. To count the number of horses in a barn, we can count the number of hoofs and then divide by 4

  33. If a finite set A has a k-to-1 correspondence to finite set B, then |B| = |A|/k

  34. I own 3 beanies and 2 ties. How many different ways can I dress up in a beanie and a tie?

  35. A Restaurant Has a Menu With5 Appetizers, 6 Entrees, 3 Salads, and 7 Desserts How many items on the menu? 5 + 6 + 3 + 7 = 21 How many ways to choose a complete meal? 5 × 6 × 3 × 7 = 630 How many ways to order a meal if I am allowed to skip some (or all) of the courses? • 6 × 7 × 4 × 8 = 1344

  36. Hobson’s Restaurant Has Only 1 Appetizer, 1 Entree, 1 Salad, and 1 Dessert 24 ways to order a meal if I might not have some of the courses Same as number of subsets of the set{Appetizer, Entrée, Salad, Dessert}

  37. 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Choice Tree For 2n n-bit Sequences We can use a “choice tree” to represent the construction of objects of the desired type

  38. 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Choice Tree For 2n n-bit Sequences 011 101 111 010 100 110 001 000 Label each leaf with the object constructed by the choices along the path to the leaf

  39. 0 1 0 1 0 1 0 1 0 1 0 1 0 1 2 choices for first bit× 2 choices for second bit× 2 choices for third bit : :× 2 choices for the nth

  40. Leaf Counting Lemma Let T be a depth-n tree when each node at depth 0  i  n-1 has Pi+1 children The number of leaves of T is given by: P1P2…Pn

  41. Choice Tree A choice tree is a rooted, directed tree with an object called a “choice” associated with each edge and a label on each leaf

  42. A choice tree provides a “choice tree representation” of a set S, if 1. Each leaf label is in S, and each element of S is some leaf label 2. No two leaf labels are the same

  43. We will now combine the correspondence principle with the leaf counting lemma to make a powerful counting rule for choice tree representation.

  44. Product Rule IF set S has a choice tree representation with P1 possibilities for the first choice, P2 for the second, P3 for the third, and so on, THEN there are P1P2P3…Pn objects in S Proof: There are P1P2P3…Pn leaves of the choice tree which are in 1-1 onto correspondence with the elements of S.

  45. Product Rule (Rephrased) Suppose every object of a set S can be constructed by a sequence of choices with P1 possibilities for the first choice, P2 for the second, and so on. IF 1. Each sequence of choices constructs an object of type S AND 2. No two different sequences create thesame object THEN There are P1P2P3…Pn objects of type S

  46. How Many Different Orderings of a Deck With 52 Cards? What object are we making? Ordering of a deck Construct an ordering of a deck by a sequence of 52 choices: 52 possible choices for the first card; 51 possible choices for the second card; : : 1 possible choice for the 52nd card. By product rule: 52 × 51 × 50 × … × 2 × 1 = 52!

  47. A permutation or arrangement of n objects is an ordering of the objects The number of permutations of n distinct objects is n!

  48. How many sequences of 7 letters are there? 267 (26 choices for each of the 7 positions)

  49. How many sequences of 7 letters contain at least two of the same letter? 267 - 26×25×24×23×22×21×20 number of sequences containing all different letters

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